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If a ball is thrown into the air with a velocity of 46 ft/s, its height in feet t seconds later is given by y = 46t − 16t2.

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you want maximum height reached?

[tex]\frac{dy}{dt} =46-32t\\\\at max. height velocity=0\\ 0=46-32t\\32 t=46\\t=46/32=23/16\\y=t(46-16t)\\ at t=\frac{23}{16} \\ y=\frac{23}{16}(46-16*\frac{23}{16} )\\\\y=\frac{529}{16} ft[/tex]

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